Connect with experts and get insightful answers to your questions on IDNLearn.com. Our platform offers reliable and comprehensive answers to help you make informed decisions quickly and easily.
Sagot :
Let's solve each equation step-by-step.
### For Problem 23:
We are given the equation:
[tex]\[ 6(p^2 - 1) = 5p \][/tex]
Step 1: Expand the equation.
[tex]\[ 6(p^2 - 1) = 6p^2 - 6 \][/tex]
Step 2: Substitute this into the original equation:
[tex]\[ 6p^2 - 6 = 5p \][/tex]
Step 3: Move all terms to one side to form a standard quadratic equation:
[tex]\[ 6p^2 - 5p - 6 = 0 \][/tex]
Step 4: Solve the quadratic equation [tex]\( 6p^2 - 5p - 6 = 0 \)[/tex]. We use the quadratic formula [tex]\( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 6 \)[/tex], [tex]\( b = -5 \)[/tex], and [tex]\( c = -6 \)[/tex].
Step 4a: Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = (-5)^2 - 4 \cdot 6 \cdot (-6) = 25 + 144 = 169 \][/tex]
Step 4b: Calculate the roots using the quadratic formula:
[tex]\[ p = \frac{-(-5) \pm \sqrt{169}}{2 \cdot 6} = \frac{5 \pm 13}{12} \][/tex]
So, we get two solutions:
[tex]\[ p = \frac{5 + 13}{12} = \frac{18}{12} = \frac{3}{2} \][/tex]
[tex]\[ p = \frac{5 - 13}{12} = \frac{-8}{12} = \frac{-2}{3} \][/tex]
Thus, the solutions for [tex]\( p \)[/tex] are:
[tex]\[ p = \frac{3}{2} \quad \text{and} \quad p = -\frac{2}{3} \][/tex]
### For Problem 27:
We are given the equation:
[tex]\[ \frac{4(x - 2)}{x - 3} + \frac{3}{x} = \frac{-3}{x(x - 3)} \][/tex]
Step 1: Multiply through by [tex]\( x(x - 3) \)[/tex] to eliminate denominators:
[tex]\[ 4(x - 2) \cdot x + 3(x - 3) = -3 \][/tex]
Step 2: Expand and simplify the equation:
[tex]\[ 4x(x - 2) + 3x - 9 = -3 \][/tex]
[tex]\[ 4x^2 - 8x + 3x - 9 = -3 \][/tex]
[tex]\[ 4x^2 - 5x - 9 = -3 \][/tex]
Step 3: Move all terms to one side to form a standard quadratic equation:
[tex]\[ 4x^2 - 5x - 9 + 3 = 0 \][/tex]
[tex]\[ 4x^2 - 5x - 6 = 0 \][/tex]
Step 4: Solve the quadratic equation [tex]\( 4x^2 - 5x - 6 = 0 \)[/tex]. We use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 4 \)[/tex], [tex]\( b = -5 \)[/tex], and [tex]\( c = -6 \)[/tex].
Step 4a: Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = (-5)^2 - 4 \cdot 4 \cdot (-6) = 25 + 96 = 121 \][/tex]
Step 4b: Calculate the roots using the quadratic formula:
[tex]\[ x = \frac{-(-5) \pm \sqrt{121}}{2 \cdot 4} = \frac{5 \pm 11}{8} \][/tex]
So, we get two solutions:
[tex]\[ x = \frac{5 + 11}{8} = \frac{16}{8} = 2 \][/tex]
[tex]\[ x = \frac{5 - 11}{8} = \frac{-6}{8} = \frac{-3}{4} \][/tex]
Thus, the solutions for [tex]\( x \)[/tex] are:
[tex]\[ x = 2 \quad \text{and} \quad x = -\frac{3}{4} \][/tex]
### Summary:
The solutions are:
- For [tex]\( 6(p^2 - 1) = 5p \)[/tex]: [tex]\( p = \frac{3}{2} \)[/tex] and [tex]\( p = -\frac{2}{3} \)[/tex]
- For [tex]\( \frac{4(x-2)}{x-3} + \frac{3}{x} = \frac{-3}{x(x-3)} \)[/tex]: [tex]\( x = 2 \)[/tex] and [tex]\( x = -\frac{3}{4} \)[/tex]
### For Problem 23:
We are given the equation:
[tex]\[ 6(p^2 - 1) = 5p \][/tex]
Step 1: Expand the equation.
[tex]\[ 6(p^2 - 1) = 6p^2 - 6 \][/tex]
Step 2: Substitute this into the original equation:
[tex]\[ 6p^2 - 6 = 5p \][/tex]
Step 3: Move all terms to one side to form a standard quadratic equation:
[tex]\[ 6p^2 - 5p - 6 = 0 \][/tex]
Step 4: Solve the quadratic equation [tex]\( 6p^2 - 5p - 6 = 0 \)[/tex]. We use the quadratic formula [tex]\( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 6 \)[/tex], [tex]\( b = -5 \)[/tex], and [tex]\( c = -6 \)[/tex].
Step 4a: Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = (-5)^2 - 4 \cdot 6 \cdot (-6) = 25 + 144 = 169 \][/tex]
Step 4b: Calculate the roots using the quadratic formula:
[tex]\[ p = \frac{-(-5) \pm \sqrt{169}}{2 \cdot 6} = \frac{5 \pm 13}{12} \][/tex]
So, we get two solutions:
[tex]\[ p = \frac{5 + 13}{12} = \frac{18}{12} = \frac{3}{2} \][/tex]
[tex]\[ p = \frac{5 - 13}{12} = \frac{-8}{12} = \frac{-2}{3} \][/tex]
Thus, the solutions for [tex]\( p \)[/tex] are:
[tex]\[ p = \frac{3}{2} \quad \text{and} \quad p = -\frac{2}{3} \][/tex]
### For Problem 27:
We are given the equation:
[tex]\[ \frac{4(x - 2)}{x - 3} + \frac{3}{x} = \frac{-3}{x(x - 3)} \][/tex]
Step 1: Multiply through by [tex]\( x(x - 3) \)[/tex] to eliminate denominators:
[tex]\[ 4(x - 2) \cdot x + 3(x - 3) = -3 \][/tex]
Step 2: Expand and simplify the equation:
[tex]\[ 4x(x - 2) + 3x - 9 = -3 \][/tex]
[tex]\[ 4x^2 - 8x + 3x - 9 = -3 \][/tex]
[tex]\[ 4x^2 - 5x - 9 = -3 \][/tex]
Step 3: Move all terms to one side to form a standard quadratic equation:
[tex]\[ 4x^2 - 5x - 9 + 3 = 0 \][/tex]
[tex]\[ 4x^2 - 5x - 6 = 0 \][/tex]
Step 4: Solve the quadratic equation [tex]\( 4x^2 - 5x - 6 = 0 \)[/tex]. We use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 4 \)[/tex], [tex]\( b = -5 \)[/tex], and [tex]\( c = -6 \)[/tex].
Step 4a: Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = (-5)^2 - 4 \cdot 4 \cdot (-6) = 25 + 96 = 121 \][/tex]
Step 4b: Calculate the roots using the quadratic formula:
[tex]\[ x = \frac{-(-5) \pm \sqrt{121}}{2 \cdot 4} = \frac{5 \pm 11}{8} \][/tex]
So, we get two solutions:
[tex]\[ x = \frac{5 + 11}{8} = \frac{16}{8} = 2 \][/tex]
[tex]\[ x = \frac{5 - 11}{8} = \frac{-6}{8} = \frac{-3}{4} \][/tex]
Thus, the solutions for [tex]\( x \)[/tex] are:
[tex]\[ x = 2 \quad \text{and} \quad x = -\frac{3}{4} \][/tex]
### Summary:
The solutions are:
- For [tex]\( 6(p^2 - 1) = 5p \)[/tex]: [tex]\( p = \frac{3}{2} \)[/tex] and [tex]\( p = -\frac{2}{3} \)[/tex]
- For [tex]\( \frac{4(x-2)}{x-3} + \frac{3}{x} = \frac{-3}{x(x-3)} \)[/tex]: [tex]\( x = 2 \)[/tex] and [tex]\( x = -\frac{3}{4} \)[/tex]
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Accurate answers are just a click away at IDNLearn.com. Thanks for stopping by, and come back for more reliable solutions.
